Let $M$ be a $n$ dimensional smooth manifold (with a smooth altas). For any $p\in M$, let \begin{align} C^\infty(p)=\Big\{ f:\,M\longrightarrow\mathbb R\,\big|\,f\text{ smooth at }p \Big\}. \end{align} Given a smooth curve $\,c:\,(-\varepsilon,\varepsilon)\longrightarrow M\,$ and $\,p=c(0)$. We define a functional \begin{align} c'(0):\ C^\infty(p)\,&\longrightarrow\,\mathbb R \\ f\,&\longmapsto\, c(0)[f]=\frac{d(f\circ c)}{dt}(0). \end{align} Let $U\subseteq M$ be an open neighborhood of $p$ and a chart $\,\varphi^{-1}:\,U\longrightarrow\mathbb R^n$. Let $$\varphi^{-1}\circ c=(x_1,\dots,x_n)=x:\ (-\varepsilon,\varepsilon)\longrightarrow M.$$ Then we have \begin{align} c'(0)[f]&=\frac{d(f\circ\varphi\circ\varphi^{-1}\circ c)}{dt}(0) \\ &=\frac{d(f\circ\varphi\circ x)}{dt}(0) \\ &=\frac{\partial (f\circ\varphi)}{\partial x_1}x(0)\frac{dx_1}{dt}(0)+\cdots+\frac{\partial (f\circ\varphi)}{\partial x_n}x(0)\frac{dx_n}{dt}(0). \end{align} A tangent vector of $M$ at $p$ is defined as a functional $c'(0)$ for some smooth curve $c$ in $M$, and the set of all tangent vector at $p$ is denoted by $TM_p$.
So here's the thing. This definition is fine to me, if $M$ is considered to be a manifold. For some reasons, I really don't like to use this definition when $M$ is a vector space over $\mathbb R$. My opinion is, since $M$ is a vector space, it is better to use tangent vector as a vector in $M$, while considering a tangent vector as a "tangent functional" can make the problem more complicated than needed.
For example, a smooth vector field on $M$ is a smooth map $\,X:\,M\longrightarrow TM,\ p\longmapsto (p,X_p),\ X_p\in TM_p$. So a basis for $TM_p$ is given by these derrivative $\displaystyle\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_n}$, where each of these is a map \begin{align} \frac{\partial}{\partial x_i}:\ C^\infty(p)&\longrightarrow\mathbb R \\ f&\longmapsto \frac{\partial f}{\partial x_i}\,=\,\frac{\partial (f\circ\varphi)}{\partial x_1}x(0)\frac{dx_1}{dt}(0), \end{align} which is complicated enough for a simply case as a vector space.
Therefore, I want to have another definition for the tangent vector which apply for vector space only, such that each tangent vector is also a vector of that space. This is my solution :
Since $M$ is a vector space, there exists a linear isomorphism $T:\,M\longrightarrow\mathbb R^n$, which is also a diffeomorphism. Now, let $N$ be a $m$-smooth submanifold of $M$, then $T(N)\subseteq\mathbb R^n$ is also a $m$-smooth manifold with parametrization $\varphi:\,\mathbb R^m\longrightarrow T(N)$. At each point $T(p)\in T(N)$, a tangent vector is given by $$v\,=\,J_{\varphi}(T(p))\cdot [u]\in\mathbb R^n,\ u\in\mathbb R^m. $$ Hence a tangent vector at $p\in N$ can be consider as $$T^{-1}(v)\,=\, T^{-1}\big[J_{\varphi}(T(p))\cdot [u]\big]\in N. $$ If we only interest in the coordinate of vector, then consider the standard basis $\mathcal B$ of $\mathbb R^n$. The representation matrix of $T$ is $\big[T\big]_{T^{-1}(\mathcal B),\mathcal B}=I_n$. Therefore \begin{align} \big[T^{-1}(v) \big]_{T^{-1}(\mathcal B)}\,=\,J_{\varphi}(T(p))\cdot[u]\,=\,[v]. \end{align}
What do you think of my approach ?